Central Forces: Spring-2025
HW 07 (SOLUTION): Due W4 D2
- Working with Representations on the Ring
S1 5244S
The following are 3 different representations for the \(\textbf{same}\) state on a quantum ring for \(r_0=1\) \begin{equation} \left|{\Phi_a}\right\rangle = i\sqrt{\frac{ 2}{12}}\left|{3}\right\rangle - \sqrt{\frac{ 1}{12}}\left|{1}\right\rangle +\sqrt{\frac{ 3}{12}}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i\sqrt{\frac{ 2}{ 12}}\left|{-1}\right\rangle +\sqrt{\frac{ 4}{12}}\left|{-3}\right\rangle \end{equation} \begin{equation} \left| \Phi_b\right\rangle \doteq \left( \begin{matrix} \vdots \\ i\sqrt{\frac{ 2}{12}}\\ 0 \\ -\sqrt{\frac{ 1}{12}} \\ \sqrt{\frac{ 3}{12}}e^{i\frac{\pi}{4}} \\ -i\sqrt{\frac{ 2}{12}}\\ 0 \\ \sqrt{\frac{4}{12} }\\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0 \end{matrix} \end{equation} \begin{equation} \Phi_c(\phi) \doteq \sqrt{\frac{1}{24 \pi}} \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) \end{equation}
With each representation of the state given above, explicitly calculate the probability that \(L_z=-1\hbar\). Then, calculate all other non-zero probabilities for values of \(L_z\) with a method/representation of your choice.
We always calculate the probabilties as the norm squared of the inner product of our given state on the basis state corresponding to the desired measurement. For \(L_z=-1\hbar\), the desired state is \(\left|{-1}\right\rangle \): \[{\cal{P}}(L_z=-1\hbar)=\left|\left\langle {-1}\middle|{\Phi_a}\right\rangle \right|^2\] \[=\left|\left\langle {-1}\right|\left(i\sqrt\frac{ 2}{12}\left|{3}\right\rangle - \sqrt\frac{ 1}{12}\left|{1}\right\rangle +\sqrt\frac{ 3}{12}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i\sqrt\frac{ 2}{ 12}\left|{-1}\right\rangle +\sqrt\frac{ 4}{12}\left|{-3}\right\rangle \right) \right|^2\] Foiling this out, only the inner product of the ket \(\left|{-1}\right\rangle \) isn't orthogonal to \(\left\langle {-1}\right|\) and so only that inner product yields a non-zero result: \[{\cal{P}}(L_z=-1\hbar)=\left|-i\sqrt\frac{ 2}{ 12}\left\langle {-1}\middle|{-1}\right\rangle \right|^2\] \[{\cal{P}}(L_z=-1\hbar)=\left|-i\sqrt\frac{ 2}{ 12} \right|^2=\frac{2}{12}=\frac{1}{6}\]
For the matrix notation: \begin{equation} \left|{-1}\right\rangle \doteq \left( \begin{matrix} \vdots \\ 0 \\0 \\0 \\0\\1 \\0 \\0 \\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0\end{matrix} \end{equation} We need the bra version of this vector which is the Hermitian Adjoint (the complex conjugate and transpose) \begin{equation} \left\langle {-1}\right|\doteq \stackrel{\stackrel{m=0}{\downarrow}} {\left( \begin{matrix} \cdots &0 &0 &0 &0 &1 &0 &0 &\cdots \end{matrix}\right)} \end{equation} We can now perform the needed matrix multiplication \begin{align} \left\langle {-1}\middle|{\Phi_b}\right\rangle &= \left( \begin{matrix} \cdots &0 &0 &0 &0 &1 &0 &0 &\cdots \end{matrix}\right) \left( \begin{matrix} \vdots \\ i\sqrt\frac{ 2}{12}\\ 0 \\ -\sqrt\frac{ 1}{12} \\ \sqrt\frac{ 3}{12}e^{i\frac{\pi}{4}} \\ -i\sqrt\frac{ 2}{12}\\ 0 \\ \sqrt\frac{4}{12} \\ \vdots \end{matrix}\right) \begin{matrix} \leftarrow m=0\end{matrix}\\ &= -i\sqrt{\frac{2}{12}} \end{align} And take the norm squared to get the probability \begin{equation} {\cal{P}}(L_z=-1\hbar)= \frac{2}{12}=\frac{1}{6} \end{equation}
Finally for the wavefunction representation of the eigenstates, we can find the inner product: \begin{align} \left\langle {-1}\middle|{\Phi_c}\right\rangle &=\int_0^{2\pi}\Phi_{-1}^*(\phi)\Phi_c(\phi)~d\phi \\&=\int_0^{2\pi}\left(\frac{1}{\sqrt{2\pi}}e^{-i\phi}\right)^* \sqrt {\frac{1}{24 \pi}} \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) ~d\phi \\&=\frac{1}{2\pi}\sqrt{\frac{1}{12}}\int_0^{2\pi}\left(e^{i\phi}\right) \left( i\sqrt{2}e^{i 3 \phi} -e^{i\phi} +\sqrt{3}e^{i\frac{\pi}{4}} -i \sqrt{2} e^{-i\phi} + \sqrt{4}e^{-i 3 \phi} \right) ~d\phi \\&=\frac{1}{2\pi}\sqrt{\frac{1}{12}}\int_0^{2\pi} \left( i\sqrt{2}e^{i 4 \phi} -e^{i2\phi} +\sqrt{3}e^{i\frac{\pi}{4}}e^{i\phi} -i \sqrt{2} + \sqrt{4}e^{-i 2 \phi} \right) ~d\phi \end{align} This looks like a sum of Kronecker deltas. \begin{equation} \delta_{nm}=\frac{1}{2\pi}\int_0^{2\pi}e^{i(n-m)\phi}d\phi \end{equation} This means that all terms of this integration will be zero, except the term with the \(0\) in the exponent. \begin{equation} \left\langle {-1}\middle|{\Phi_c}\right\rangle =\frac{1}{2\pi}\sqrt{\frac{1}{12}}\int_0^{2\pi}-i \sqrt{2}d\phi = \frac{-i}{2\pi}\sqrt{\frac{1}{6}}\int_0^{2\pi}d\phi=-i\sqrt{\frac{1}{6}} \end{equation} We take the norm squared of this inner product to get the probability, so: \[\cal{P}(L_z=-1\hbar)=\frac{1}{6}\] Through any of the representation, we can find the remaining non-zero probabilities, with the easiest way being to read off the probabilities from from the ket representatio n and norm square them: \[{\cal{P}}(L_z=3\hbar)=\frac{1}{6}\] \[{\cal{P}}(L_z=1\hbar)=\frac{1}{12}\] \[{\cal{P}}(L_z=0\hbar)=\frac{1}{4}\] \[{\cal{P}}(L_z=-1\hbar)=\frac{1}{6}\] \[{\cal{P}}(L_z=-3\hbar)=\frac{1}{3}\]
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Explain how you could be sure you calculated all of the non-zero probabilities.
The easiest way would be to check if the sum of all the probabilities add into 1, and they do. Or you can argue that only kets represented in the state can be thought of as having non-zero probabilities of measurement.
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If you measured the \(z\)-component of angular momentum to be \(3\hbar\), what would the
state of the particle be immediately after the measurement is made?
It will be in the \(\vert 3\rangle\) state, since the function is unambigiously in that state based on our measurment (this simple case will always happen when non-degenerate values are measured).
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With each representation of the state given above, explicitly calculate the probability that \(E=\frac{9}{2}\frac{\hbar^2}{I}\). Then, calculate all other non-zero probabilities for values of \(E\) with a method of your choice.
The big thing to remember here is that energies on the ring can be degenerate, which means we need to take the probabilities of being in the \(\left|{+m}\right\rangle \) and \(\left|{-m}\right\rangle \) and add them together in order to get the total probability. Like so: \[{\cal{P}}\left(E=\frac{m^2\hbar^2}{2I}\right)=\left|\left\langle {+m}\middle|{\Phi}\right\rangle \right|^2+\left|\left\langle {-m}\middle|{\Phi}\right\rangle \right|^2\] The individual inner products and probabilities will be exactly the same as for \(L_z\) we just need to add them together properly to take account for the degeneracy in the energies: \[{\cal{P}}\left(E=\frac{9}{2}\frac{\hbar^2}{I}\right)=\left|\left\langle {3}\middle|{\Phi}\right\rangle \right|^2+\left|\left\langle {-3}\middle|{\Phi}\right\rangle \right|^2=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\] \[{\cal{P}}\left(E=\frac{1}{2}\frac{\hbar^2}{I}\right)=\left|\left\langle {1}\middle|{\Phi}\right\rangle \right|^2+\left|\left\langle {-1}\middle|{\Phi}\right\rangle \right|^2=\frac{1}{12}+\frac{1}{6}=\frac{1}{4}\] \[{\cal{P}}\left(E=0\frac{\hbar^2}{I}\right)=\left|\left\langle {0}\middle|{\Phi}\right\rangle \right|^2=\frac{1}{4}\]
If you measured the energy of the state to be \(\frac{9}{2}\frac{\hbar^2}{I}\), what would the state of the particle be immediately after the measurement is made?
Because the state corresponding to this energy is degenerate, we need to consider the a superposition of states coming out, and use the projection postulate! \[\left|{\Psi}\right\rangle = \frac{P_n\left|{\Phi}\right\rangle }{\sqrt{\langle{\Phi}|P_n|{\Phi}\rangle}},\] where \(P_n\) is the projection operator for the subset of eigenstates associated with \(\frac{9}{2}\frac{\hbar^2}{I}\) and \(\left|{\Psi}\right\rangle \) is the new state after the measurement. Therefore, our projection operator is:
\[P_n = \left|{3}\right\rangle \left\langle {3}\right| + \left|{-3}\right\rangle \left\langle {-3}\right|.\]
I'll do the calculation in pieces so that it fits on the page: \begin{align} P_n\left|{\Phi}\right\rangle &=\left(\left|{3}\right\rangle \left\langle {3}\right| + \left|{-3}\right\rangle \left\langle {-3}\right|\right) \left(i{\scriptstyle\sqrt{\frac{ 2}{12}}}\left|{3}\right\rangle - {\scriptstyle\sqrt{\frac{ 1}{12}}}\left|{1}\right\rangle +{\scriptstyle\sqrt{\frac{ 3}{12}}}e^{i\frac{\pi}{4}}\left|{0}\right\rangle -i{\scriptstyle\sqrt{\frac{ 2}{ 12}}}\left|{-1}\right\rangle +{\scriptstyle\sqrt{\frac{ 4}{12}}}\left|{-3}\right\rangle \right)\\ &=\left(\left|{3}\right\rangle i{\scriptstyle\sqrt{\frac{ 2}{12}}} +\left|{-3}\right\rangle {\scriptstyle\sqrt{\frac{ 4}{12}}} \right) \end{align} and \begin{align} \left\langle {\Phi}\right|P_n\left|{\Phi}\right\rangle &=\left(-i{\scriptstyle\sqrt{\frac{ 2}{12}}}\left\langle {3}\right| - {\scriptstyle\sqrt{\frac{ 1}{12}}}\left\langle {1}\right| +{\scriptstyle\sqrt{\frac{ 3}{12}}}e^{-i\frac{\pi}{4}}\left\langle {0}\right| +i{\scriptstyle\sqrt{\frac{ 2}{ 12}}}\left\langle {-1}\right| +{\scriptstyle\sqrt{\frac{ 4}{12}}}\left\langle {-3}\right| \right) \left(\left|{3}\right\rangle i{\scriptstyle\sqrt{\frac{ 2}{12}}} +\left|{-3}\right\rangle {\scriptstyle\sqrt{\frac{ 4}{12}}} \right)\\ &= (-i{\scriptstyle\sqrt{\frac{ 2}{12}}})(i{\scriptstyle\sqrt{\frac{ 2}{12}}}) +({\scriptstyle\sqrt{\frac{ 4}{12}}})({\scriptstyle\sqrt{\frac{ 4}{12}}})\\ &=\frac{1}{2} \end{align}
Thus, the state of the particle immediately after the measurement of \(\displaystyle E = \frac{9}{2}\frac{\hbar^2}{I}\) is: \begin{align} \left|{\Psi}\right\rangle &= \frac{P_n\left|{\Phi}\right\rangle }{\sqrt{\langle{\Phi}|P_n|{\Phi}\rangle}}\\ &=\frac{\left|{3}\right\rangle i{\scriptstyle\sqrt{\frac{ 2}{12}}} +\left|{-3}\right\rangle {\scriptstyle\sqrt{\frac{ 4}{12}}}} {\sqrt{\frac{1}{2}}}\\ &=i\sqrt{\frac{1}{3}}\left|{3}\right\rangle +\sqrt{\frac{2}{3}}\left|{-3}\right\rangle \end{align}
Sensemaking: Notice that this state is normalized, as it must be, i.e. the sum of the squares of the norms of the coefficients is equal to one.
- Ring Function
S1 5244S
Consider the normalized wavefunction \(\Phi\left(\phi\right)\) for a
quantum mechanical particle of mass \(\mu\) constrained to move on a
circle of radius \(r_0\), given by:
\begin{equation}
\Phi\left(\phi\right)= \frac{N}{2+\cos(3\phi)}
\end{equation}
where \(N\) is the normalization constant.
Find \(N\).
See the attached Mathematica notebook for the value of the integrals in this problem and for the code for the plots.
To find \(N\), normalize the wave function. \begin{align} \int_{0}^{2\pi} \left|\Phi\right|^2\,r_0\,d\phi &= 1\\ |N|^2\,r_0 \int_{0}^{2\pi} \frac{1}{(2+\cos{3\phi})^2}\,d\phi &= 1\\ |N|^2\,r_0 \left(\frac{4\sqrt{3}\pi}{9}\right) &= 1\label{norm}\\ N &= \frac{3^{3/4}}{2\sqrt{\pi r_0}} \end{align} where in Eqn. (\ref{norm}) we have evaluated the integral using Mathematica.
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Plot this wave function.
The unnormalized wavefunction is \(\frac{1}{2+\cos{3\phi}}\)
A plot of the wave function \(\Phi(\phi)\). - Plot the probability density.
The probability density is \(\left\vert \frac{1}{2+3\cos\phi}\right\vert^2\).
A plot of the probability density \(\vert\Phi(\phi)\vert^2\). - Find the probability that if you measured \(L_z\) you would get \(3\hbar\).
\begin{align} \mathcal{P}(L_z=3\hbar)&=\left\vert\left\langle {3}\middle|{\Phi}\right\rangle \right\vert^2\\ &=\left\vert\int_0^{2\pi} \left(\frac{1}{\sqrt{2\pi r_0}}e^{-3i\phi}\right) \left(\frac{N}{2+\cos(3\phi)}\right)\, r_0\, d\phi \right\vert^2\\ &=\left\vert\int_0^{2\pi} \left(\frac{1}{\sqrt{2\pi r_0}}(\cos{3\phi}-i\sin{3\phi})\right) \left(\frac{N}{2+\cos(3\phi)}\right)\, r_0\, d\phi \right\vert^2\\ &=\frac{\vert N\vert^2\, r_0}{2\pi}\left\vert\int_0^{2\pi} \left(\left(\frac{\cos{3\phi}}{2+\cos(3\phi)}\right) -\left(\frac{\sin{3\phi}}{2+\cos(3\phi)}\right)\right)\, d\phi \right\vert^2\label{Lz}\\ &=\frac{\vert N\vert^2\, r_0}{2\pi} \left\vert\left(2-\frac{4}{\sqrt{3}}\right)\pi+0\right\vert^2\\ &=\frac{\sqrt{3}}{2}\left(\sqrt{3}-2\right)^2 \end{align} where in Eqn. (\ref{Lz}) we have evaluated the integral using Mathematica. See the attached notebook.
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What is the expectation value of \(L_z\) in this state?
In the position representation, \(L_z=-i\hbar \frac{d}{d\phi}\) is a differential operator. Make sure to sandwich it between the "bra" and "ket" and only let it operate on the "ket." \begin{align} \langle L_z \rangle &= \int_{0}^{2\pi}\Phi^*\hat{L_z}\Phi\,r_0\,d\phi\\ &= \int_{0}^{2\pi} \frac{N}{2+\cos{3\phi}} \; \left(-i\hbar \frac{d}{d \phi}\right) \frac{N}{2+\cos{3\phi}}\,r_0\,d\phi\\ &= -i\hbar |N|^2 \int_{0}^{2\pi} \left(\frac{1}{2+\cos{3\phi}} \right) \left(\frac{3\sin{3\phi}}{(2+\cos{3\phi})^2}\right) \,d\phi \label{expect}\\ &= 0 \end{align} where in Eqn. (\ref{expect}) we have evaluated the integral using Mathematica. See the attached notebook.
- Ring Table
S1 5244S
Attached, you will find a table showing different representations of physical
quantities associated with a quantum particle confined to a ring. Fill in all of the missing entries. Hint: You may look ahead. We filled out a number of the entries throughout the table to give you hints about what the forms of the other entries might be.
pdf link for the Table
or doc link for the Table
pdf link for the completed Table
- Confidence Rating
S1 5244S
After solving each problem on the assignment, indicate your answers to the following questions for each problem. Answer for the problem as a whole, even if the problem has multiple parts.
- Question Confidence How confident are you that you are interpreting the problem the way the instructor intends?
For the rest of the questions, assume you have interpreted the problem correctly - Problem Confidence How confident are you that you could independentl come up with a correct solution process to a similar problem on a future problem set?
- Answer Confidence How confident are you that your final answer to this question is correct (not solution process)?
- Makes Sense To what degree do you understand how your answer fits (or does not fit) the physical or mathematical situation of the problem?
- Question Confidence How confident are you that you are interpreting the problem the way the instructor intends?